Question #209785

Assume that A vector a of length ||a|| =3 units. In addition, a points in a direction that is 135° counterclockwise from the positive x-axis, and a vector b in the xy-plane has a length ||b|| =1/3 and points in the positive y-direction.

a.) Find a.b

b.) calculate the distance between the point(-1,3) and the line 2x - 2y-5=0


1
Expert's answer
2021-06-23T14:16:07-0400

a)


a=3cos135°,3sin135°,\vec a=\langle3\cos135\degree, 3\sin 135\degree\rangle,

b=0,13\vec b=\langle0,\dfrac{1}{3}\rangle

ab=3(22)(0)+3(22)(13)=22\vec a\cdot \vec b=3(-\dfrac{\sqrt{2}}{2})(0)+3(\dfrac{\sqrt{2}}{2})(\dfrac{1}{3})=\dfrac{\sqrt{2}}{2}

Or

Angle between two vectors (a,b)=135°90°=45°\angle(\vec a, \vec b)=135\degree-90\degree =45\degree

ab=abcos((ab))\vec a\cdot \vec b=|\vec a||\vec b|\cos(\angle(\vec a\cdot \vec b))

=3(13)cos45°=22=3(\dfrac{1}{3})\cos 45\degree=\dfrac{\sqrt{2}}{2}



ab=22\vec a\cdot \vec b=\dfrac{\sqrt{2}}{2}


b)


d=Ax0+By0+CA2+B2d=\dfrac{|Ax_0+By_0+C|}{\sqrt{A^2+B^2}}

Point (1,3):x0=1,y0=3(-1, 3): x_0=-1, y_0=3

Line 2x2y5=02x - 2y-5=0


d=2(1)2(3)5(2)2+(2)2=1324d=\dfrac{|2(-1)-2(3)-5|}{\sqrt{(2)^2+(-2)^2}}=\dfrac{13\sqrt{2}}{4}

The distance is 1324.\dfrac{13\sqrt{2}}{4}.



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